
The amount of charge required to liberate 9 gm of aluminum (atomic weight = 27 and valency = 3) in the process of electrolysis is (Faraday’s number = 965000 coulombs/gm equivalent)
A) 321660 coulombs
B) 69500 coulombs
C) 289500 coulombs
D) 96500 coulombs
Answer
233.4k+ views
Hint: The problem is from the chemical effect of the current section of physics. We have to use Faraday's first law of electrolysis to solve this problem.
Formula Used:Faraday's first law of electrolysis:
\(m = zq\)
Where m = mass of a substance liberated or deposited at an electrode, q = charge and z = electrochemical equivalent of the substance.
\(z = \frac{{Atomic{\rm{ }}mass}}{{Valency}} \times \frac{1}{{96500}}gm{C^{ - 1}}\)
Complete step by step solution:The given mass of aluminum = 9 gm
electrochemical equivalent of aluminum, \(z = \frac{{Atomic{\rm{ }}mass}}{{Valency}} \times \frac{1}{{96500}}gm{C^{ - 1}}\)
\(z = \frac{{27}}{3} \times \frac{1}{{96500}}\)
\(z = 9.3264 \times {10^{ - 5}}gm{C^{ - 1}}\)
Using Faraday's first law of electrolysis, we can find the amount charged.
\(m = zq\)
\(q = \frac{m}{z} = \frac{9}{{9.3264 \times {{10}^{ - 5}}}}\)
\(q = 96500C\)
Hence, the correct option is Option (D).
Additional Information:The mass of the substance released or deposited on an electrode during electrolysis is directly proportional to the amount of electric charge carried through the electrolyte, according to Faraday's first law.
If m = mass of a substance liberated or deposited at an electrode and q = charge. Then according to Faraday's law of electrolysis,
\(m \propto q \Rightarrow m = zq\)
Where z = electrochemical equivalent of the substance.
According to Faraday's second law of electrolysis, the masses of ions deposited at the electrodes are inversely proportional to their chemical equivalents when the same amount of electricity is applied to various electrolytes.
Note: The Faraday constant represents the amount of electric charge carried by one mole, or Avogadro's number of electrons. It is a crucial constant in physics, electronics, and chemistry. The measurement is given in coulombs per mole (C/mol).
Formula Used:Faraday's first law of electrolysis:
\(m = zq\)
Where m = mass of a substance liberated or deposited at an electrode, q = charge and z = electrochemical equivalent of the substance.
\(z = \frac{{Atomic{\rm{ }}mass}}{{Valency}} \times \frac{1}{{96500}}gm{C^{ - 1}}\)
Complete step by step solution:The given mass of aluminum = 9 gm
electrochemical equivalent of aluminum, \(z = \frac{{Atomic{\rm{ }}mass}}{{Valency}} \times \frac{1}{{96500}}gm{C^{ - 1}}\)
\(z = \frac{{27}}{3} \times \frac{1}{{96500}}\)
\(z = 9.3264 \times {10^{ - 5}}gm{C^{ - 1}}\)
Using Faraday's first law of electrolysis, we can find the amount charged.
\(m = zq\)
\(q = \frac{m}{z} = \frac{9}{{9.3264 \times {{10}^{ - 5}}}}\)
\(q = 96500C\)
Hence, the correct option is Option (D).
Additional Information:The mass of the substance released or deposited on an electrode during electrolysis is directly proportional to the amount of electric charge carried through the electrolyte, according to Faraday's first law.
If m = mass of a substance liberated or deposited at an electrode and q = charge. Then according to Faraday's law of electrolysis,
\(m \propto q \Rightarrow m = zq\)
Where z = electrochemical equivalent of the substance.
According to Faraday's second law of electrolysis, the masses of ions deposited at the electrodes are inversely proportional to their chemical equivalents when the same amount of electricity is applied to various electrolytes.
Note: The Faraday constant represents the amount of electric charge carried by one mole, or Avogadro's number of electrons. It is a crucial constant in physics, electronics, and chemistry. The measurement is given in coulombs per mole (C/mol).
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